# Problem Set 6 Solutions

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1 6.04/18.06J Mathmatics for Computr Scic March 15, 005 Srii Dvadas ad Eric Lhma Problm St 6 Solutios Du: Moday, March 8 at 9 PM Problm 1. Sammy th Shar is a fiacial srvic providr who offrs loas o th followig trms. Sammy loas a clit m dollars i th morig. This puts th clit m dollars i dbt to Sammy. Each vig, Sammy first chargs a srvic f, which icrass th clit s dbt by f dollars, ad th Sammy chargs itrst, which multiplis th dbt by a factor of p. For xampl, if Sammy s itrst rat wr a modst 5% pr day, th p would b (a) What is th clit s dbt at th d of th first day? Solutio. At th d of th first day, th clit ows Sammy (m + f)p = mp + fp dollars. (b) What is th clit s dbt at th d of th scod day? Solutio. ((m + f)p + f)p = mp + fp + fp (c) Writ a formula for th clit s dbt aftr d days ad fid a quivalt closd form. Solutio. Th clit s dbt aftr thr days is (((m + f)p + f)p + f)p = mp 3 + fp 3 + fp + fp. Gralizig from this pattr, th clit ows d d mp + fp dollars aftr d days. Applyig th formula for a gomtric sum givs: d+1 d p 1 mp + f 1 p 1 Problm. Fid closd form xprssios qual to th followig sums. Show your wor. =1

2 Problm St 6 (a) 9 i 7 i i=0 11 i Solutio. Split th xprssio ito two gomtric sris ad th apply th formula for th sum of a gomtric sris. i 9 i 7 i 9 7 i = 11 i i=0 i=0 i= = = (b) i=1 3 4i+5 Solutio. Taig th logarithm rducs this product to a asy sum. 3 log 3( Q 3 4i+5 = i=1 34i+5 ) i=1 P = 3 i=1 4i+5 = 3 (+1)+5 (c) 1 j 5/3 1 j 1/3 j=1 i=0 i Solutio. This farsom looig sum is a papr tigr; w just apply th formula for th sum of a gomtric sris followd by th formula for th sum of a arithmtic sris. i 1 1 j 5/3 1 = j 5/3 j 1/3 j=1 i=0 j=1 1 1 j 1/3 = j = j=1 1 ( + )( + 1) 3 1

4 4 Problm St 6 Problm 4. Us itgratio to fid lowr ad uppr bouds o th followig ifiit sum that diffr by at most 0.1. Show your wor S = To achiv this accuracy, add up th first fw trms xplicitly ad th us itgratio to boud all rmaiig trms. Solutio. Th sum of th first thr trms is: s = + + = A uppr boud o th rmaiig trms is: 1 1 dx = 3 x 3 Ad a lowr boud is: 1 1 dx = (x + 1) 4 3 Ovrall, w hav: = S Ths bouds diffr by 1/1 < 0.1. Th actual valu of th sum is π /6, though th proof is ot asy. Problm 5. A sasod MIT udrgraduat ca: Complt a problm st i days. Writ a papr i days. Ta a day road trip. Study for a xam i 1 day. Play foosball for a tir day. A day schdul is a squc of activitis that rquir a total of days. For xampl, hr ar thr possibl 7 day schduls: pst, papr, pst, foosball papr, study, foosball, pst, study road trip, road trip, road trip, study

5 Problm St 6 5 (a) Exprss th umbr of possibl day schduls usig a rcurrc quatio ad sufficit bas cass. Solutio. S(0) = 1, S(1) =. Ay schdul for > 1 days ds with o of 3 possibl day activitis or o of possibl 1 day activitis. So S() = S( 1) + 3S( ) for > 1. (b) Fid a closd form xprssio for th umbr of possibl day schduls by solvig th rcurrc. Solutio. Th charactristic polyomial for this liar homogous rcurrc is x x 3 = (x + 1)(x 3). Hc th solutio is of th form S() = a( 1) + b3. Lttig = 0, w coclud that a+b = 1, ad lttig = 1, w coclud a+3b =, so b = 3/4, a = 1/4, ad th solutio is: ( 1) S() =. 4 Problm 6. Fid a closd form xprssio for T (), which is dfid by th followig rcurrc: T (0) = 0 T (1) = 1 T () = 5T ( 1) 6T ( ) + 6 for all Solutio. Th charactristic quatio is x 5x + 6 = 0, which has roots x = ad x = 3. Thus, th homogous solutio is: T () = A + B 3 For a particular solutio, lt s first guss T () = c: c = 5c 6c + 6 c = 3 Our guss was corrct; T () = 3 is a particular solutio. Addig this to th homogous solutio givs th gral solutio: T () = A + B 3 + 3

6 6 Problm St 6 Substitutig = 0 ad = 1 givs: 0 = A + B = A + 3B + 3 Solvig this systm givs A = 7 ad B = 4. Thrfor: T () = Problm 7. Dtrmi which of ths choics Θ(), Θ( log ), Θ( ), Θ(1), Θ( ), Θ( l ), o of ths dscribs ach fuctio s asymptotic bhavior. Proofs ar ot rquird, but brifly xplai your aswrs. (a) + l + (l ) Solutio. Both > l ad > (l ) hold for all sufficitly larg. Thus, for all sufficitly larg : < + l + (l ) < + + So + l + (l ) = Θ(). (b) Solutio. Obsrv that: + 3 lim = 1 7 This mas, that for all sufficitly larg, th fractio lis, for xampl, btw, 0.99 ad 1.01 ad is thrfor Θ(1). (c) i=0 i+1 Solutio. Gomtric sums ar domiatd by thir largst trm, which is +1 = 4. This is Θ(4 ), which dos ot appar i th list providd. (d) l(!) Solutio. By Stirlig s formula:! π

7 Problm St 6 7 Taig logarithms givs: l(!) l( π ) = l( π ) + l Th first trm is tiy compard to th scod, which w ca rwrit as: l = l = Θ( l ) () 1 1 =1 Solutio. Th xprssio i parthss is always at last 1/ ad at most 1. Thus, w hav th bouds: =1 =1 =1 Sic th first xprssio ad th last ar both Θ( ), so is th o i th middl. Problm 8. A triagular umbr is a itgr of th form whr is a positiv itgr. = = ( + 1) (a) Dscrib a solutio to th four pg Towrs of Haoi puzzl with ( + 1)/ diss that rquirs T movs, whr: Solutio. T 1 = 1 T = T Mov all but th largst diss to aothr pg rcursivly. This rquirs T ( 1) movs. Mov th largst diss to aothr pg usig th thr pg stratgy. This rquirs 1 movs. Now mov all th othr diss o top of th largst diss rcursivly. This rquirs T ( 1) movs.

8 8 Problm St 6 Thus, with this stratgy, th total umbr of movs rquird to mov a stac of ( + 1)/ diss is T () = T ( 1) + 1. (b) Fid a closd form xprssio qual to T. Solutio. This is a ihomogous liar quatio. Lt s bgi by tryig to fid a particular solutio. Thr is both a xpotial trm ( ) ad a costat trm, so w might guss somthig of th form a + c: a + c = (a 1 + c) + 1 = (a + 1) + c 1 0 = + (c 1) Evidtly, th costat trm is c = 1, but th xpotial part is mor complicatd. Our rcip says w should xt try a particular solutio of th form a + b + 1: a + b + 1 = (a 1 + b( 1) 1 + 1) + 1 = (a b + 1) + b 1 Equatig th cofficits of th trms givs a = a b + 1, which implis b = 1. Thus, a + +1 is a particular solutio for all a. As log as w hav this dgr of frdom, w might as wll choos a so this solutio is cosistt with th boudary coditio T 1 = 1ad b do: a = 1 a = 1 Thrfor, th solutio to th rcurrc is T = ( 1) + 1. (c) Approximatly how may movs ar rquird to solv th four pg, dis Towrs of Haoi puzzl as a fuctio of? Assum is a triagular umbr. (For styl poits, ma corrct us of asymptotic otatio.) 1 Solutio. W hav = ( ) = + O(1). So th umbr of movs rquird is Θ( ).

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